Problem: Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}8x+y &= 1 \\ 8x-y &= -5\end{align*}$
Explanation: Begin by moving the $x$ -term in the second equation to the right side of the equation. $-y = -8x-5$ Divide both sides by $-1$ to isolate $y$ $y = {8x + 5}$ Substitute this expression for $y$ in the first equation. $8x+({8x + 5}) = 1$ $8x + 8x + 5 = 1$ Simplify by combining terms, then solve for $x$ $16x + 5 = 1$ $16x = -4$ $x = -\dfrac{1}{4}$ Substitute $-\dfrac{1}{4}$ for $x$ back into the top equation. $8( -\dfrac{1}{4})+y = 1$ $-2+y = 1$ $y = 3$ $y = 3$ The solution is $\enspace x = -\dfrac{1}{4}, \enspace y = 3$.